Methods of Proof – Page 1CSCI 1900 – Discrete Structures CSCI 1900 Discrete Structures Methods of Proof Reading: Kolman, Section 2.3

Methods of Proof – Page 2CSCI 1900 – Discrete Structures Past Experience Up to now we’ve used the following methods to write proofs: –Used direct proofs with generic elements, definitions, and given facts –Used proof by cases such as when we used truth tables

Methods of Proof – Page 3CSCI 1900 – Discrete Structures General Description of Process p  q denotes "q logically follows from p“ Implication may take the form (p 1  p 2  p 3  …  p n )  q q logically follows from p 1, p 2, p 3, …, p n

Methods of Proof – Page 4CSCI 1900 – Discrete Structures General Description (continued) The process is generally written as: p 1 p 2 p 3 : : p n  q

Methods of Proof – Page 5CSCI 1900 – Discrete Structures Components of a Proof The p i 's are called hypotheses or premises q is called the conclusion Proof shows that if all of the p i 's are true, then q has to be true If result is a tautology, then the implication p  q represents a universally correct method of reasoning and is called a rule of inference

Methods of Proof – Page 6CSCI 1900 – Discrete Structures Example of a Proof based on a Tautology If p implies q and q implies r, then p implies r p  q q  r  p  r By replacing the bar under q  r with the “  ”, the proof above becomes ((p  q)  (q  r))  (p  r) The next slide shows that this is a tautology and therefore is universally valid.

Methods of Proof – Page 7CSCI 1900 – Discrete Structures Tautology Example (continued) pqr p  qq  r(p  q)  (q  r) p  r((p  q)  (q  r))  (p  r) TTTTTTTT TTFTFFFT TFTFTFTT TFFFTFFT FTTTTTTT FTFTFFTT FFTTTTTT FFFTTTTT

Methods of Proof – Page 8CSCI 1900 – Discrete Structures Equivalences Some mathematical theorems are equivalences, i.e., p  q. The proof of such a theorem is equivalent with proving both p  q and q  p

Methods of Proof – Page 9CSCI 1900 – Discrete Structures modus ponens form (the method of asserting): p p  q  q Example: –p: a man used the toilet –q: the toilet seat is up –p  q: If a man used the toilet, the seat was left up Supported by the tautology (p  (p  q))  q

Methods of Proof – Page 10CSCI 1900 – Discrete Structures modus ponens (continued) pq (p  q)p  (p  q)(p  (p  q))  q TTTTT TFFFT FTTFT FFTFT

Methods of Proof – Page 11CSCI 1900 – Discrete Structures Invalid Conclusions from Invalid Premises Just because the format of the argument is valid does not mean that the conclusion is true. A premise may be false. For example: Acorns are money If acorns were money, no one would have to work  No one has to work Argument is valid since it is in modus ponens form Conclusion is false because premise p is false

Methods of Proof – Page 12CSCI 1900 – Discrete Structures Invalid Conclusion from Invalid Argument Sometimes, an argument that looks like modus ponens is actually not in the correct form. For example: If tuition was free, enrollment would increase Enrollment increased  Tuition is free Argument is invalid since its form is: p  q q  p

Methods of Proof – Page 13CSCI 1900 – Discrete Structures Invalid Argument (continued) Truth table shows that this is not a tautology: pq (p  q)(p  q)  q((p  q)  q)  p TTTTT TFFFT FTTTF FFTFT

Methods of Proof – Page 14CSCI 1900 – Discrete Structures Indirect Method Another method of proof is to use the tautology: (p  q)  (~q  ~p) The form of the proof is: ~q ~q  ~p  p

Methods of Proof – Page 15CSCI 1900 – Discrete Structures Indirect Method Example p: My address is available on a web site q: I am getting spam p  q: If my address is available on a web site, then I am getting spam ~q  ~p: If I am not getting spam, then my e- mail address must not be available on a web site This proof says that if I am not getting spam, then my address is not on a web site.

Methods of Proof – Page 16CSCI 1900 – Discrete Structures Another Indirect Method Example Prove that if the square of an integer is odd, then the integer is odd too. p: n 2 is odd q: n is odd ~q  ~p: If n is even, then n 2 is even. If n is even, then there exists an integer m for which n = 2×m. n 2 therefore would equal (2×m) 2 = 4×m 2 which must be even.

Methods of Proof – Page 17CSCI 1900 – Discrete Structures Proof by Contradiction Another method of proof is to use the tautology (p  q)  (~q)  (~p) The form of the proof is: p  q ~q  ~p

Methods of Proof – Page 18CSCI 1900 – Discrete Structures Proof by Contradiction (continued) pq (p  q) ~q (p  q)  ~q ~p (p  q)  (~q)  (~p) TTTFFFT TFFTFFT FTTFFTT FFTTTTT

Methods of Proof – Page 19CSCI 1900 – Discrete Structures Proof by Contradiction (continued) The best application for this is where you cannot possibly go through a large number (such as infinite) of cases to prove that every one is true.

Methods of Proof – Page 20CSCI 1900 – Discrete Structures Proof by Contradiction Example Prove that  (2) is irrational, i.e., cannot be represented with m/n where m and n are integers. –p:  (2) is a rational number –q: There exists integers m and n for every rational number such that the rational number can be expressed as m/n –p  q: If  (2) is a rational number, then we can find m and n –The goal is to prove that we cannot find an m and an n, i.e., ~q is true.

Methods of Proof – Page 21CSCI 1900 – Discrete Structures Proof by Contradiction Example (continued) –Assume (m/n) 2 = 2 and that m and n are in their most reduced form. This means that m 2 = 2n 2. –Therefore, m must be even and m 2 must contain 2 2 –Therefore, n must be even too. –Therefore, m/n is not in the most reduced form (we can pull a 2 out of both m and n). –This is a contradiction! Cannot come up with m and n, i.e., ~q is true –Therefore, ~p is true and  (2) must not be a rational number